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7 hosts LAN-2

Give IP –addresses to the networks below with three point-to-point networks and three local area networks. The available address space is 10.38.161.0 – 10.38.161.67 and the interface of R3 router towards LAN-3 is 10.38.161.1. 7 hosts LAN-2. 3 hosts LAN-1. R2. R2. R1. R1. R3. R3.

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7 hosts LAN-2

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  1. Give IP –addresses to the networks below with three point-to-point networks and three local area networks. The available address space is 10.38.161.0 – 10.38.161.67 and the interface of R3 router towards LAN-3 is 10.38.161.1. 7 hosts LAN-2 3 hosts LAN-1 R2 R2 R1 R1 R3 R3 10.38.161.1 17 hosts LAN-3

  2. SOLUTION: 1. Find the point-to-points networks and the local networks. 2. Define the address space needed in each network (how many interfaces?) 3. What is the size of the block (power of 2)? 4. Start allocating the IP –addresses from R3 (10.38.100.1). 5. What is the mask? 6. Remember the rule: network address/number of addresses in the network = Interger Remarks: Address 10.38.100.12/28 (or 255.255.255.240) cannot be a network address. Why not? How to calculate the network address and broadcast address?

  3. SOLUTION 1. Find the point-to-points networks and the local networks. R1-R2 R2 3 hosts R1 7 hosts LAN-1 LAN-2 R1-R3 R2-R3 R3 17 hosts LAN-3

  4. SOLUTION 2. Define the address space needed in each network (how many interfaces?) 3. What is the size of the block (power of 2)? LAN-3: 17 hosts+NWA+BCA+R3-interface=20  32 addr (5 bits) LAN-2: 7 hosts+1+1+1=10  16 addr (4 bits) LAN-1: 3 hosts+1+1+1=6  8 addr (3 bits) R1-R2: 2 hosts+1+1=4  4 addr (2 bits) R1-R3: 2+1+1=4  4 addr (2 bits) R2-R3: 2+1+1=4  4 addr (2 bits) 0 31|32 47|48 55|56 59|60 63|64 67| 32 addresses 16 addr 8 addr 4 a. 4 a. 4 a.

  5. SOLUTION:4. Start allocating the IP –addresses from R3 (10.38.100.1). 5. What is the mask? LAN-3: 10.38.100.0  10.38.100.31 (mask 27 or 255.255.255.224) 32 addr LAN-2: 10.38.100.32  10.38.100.47 (mask 28 or 255.255.255.240) 16 addr LAN-1: 10.38.100.48  10.38.100.55 (mask 29 or 255.255.255.248) 8 addr RI-R2: 10.38.100.56  10.38.100.59 (mask 30 or 255.255.255.252) 4 addr RI-R3: 10.38.100.60  10.38.100.63 (mask 30 or 255.255.255.252) 4 addr R2-R3: 10.38.100.64  10.38.100.67 (mask 30 or 255.255.255.252) 4 addr totally 68 addr 6. Remember the rule: network address/number of addresses in the network = Interger LAN-3: 0/32=0Integer R1-R2: 56/4=14Integer LAN-2: 32/16=2Integer R1-R3: 60/4=15Integer LAN-1: 48/8=6Integer R2-R3: 64:4=16Integer

  6. Address 10.38.100.12/28 (or 255.255.255.240) cannot be a network address. Why not? Mask=28  28 bits for network 32-28=4 bits for hosts16 addr NWA/addresses  12/16 NOT Integer

  7. How to calculate the network address and broadcast address? 10.38.100.12/28 = 10.38.100.12/255.255.255.240 0 0 0 0 1 1 0 0 = 12 AND 1 1 1 1 0 0 0 0 = 240 mask 0 0 0 0 0 0 0 0 = 0 NWA OR 0 0 0 0 1 1 1 1 = INV mask 0 0 0 0 1 1 1 1 = 15 BCA

  8. Suppose we think that 12 is NWA and mask is 28. There are 4 bits available for hosts =16 addr (015). 0 0 0 0 1 1 0 0 = 12 0 0 0 0 1 1 0 1 = 13 0 0 0 0 1 1 1 0 = 14 0 0 0 0 1 1 1 1 = 15 0 0 0 1 0 0 0 0 = 16  needs 5 bits etc. etc.

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